$f\,^{\prime}(x)=-27e^x$ and $f(6)=36-27e^6$. $f(0) = $
Solution: Finding $f(x)$ We have $f'(x)=-27e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-27e^x)\,dx \\\\ & = {-27e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(6)=36-27e^6$. Here's what we get when we plug in $6$ : $\begin{aligned}f(6)&={-27e^{6}} {+ C} \end{aligned}$ We are given that this must equal $36-27e^6$ : $36-27e^6 = {-27e^{6}} {+ C}$ Solving the equation gives us ${C=36}$. Finding $f(0)$ Now, we have that $f(x)={-27e^x} {+ 36}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=-27e^0 + 36\\\\ &=9 \end{aligned}$ The answer $f(0) = 9$